Question

unit: exponents and polynomials progress: the movement of the progress bar may be uneven because questions can be worth more or less (including zero) a evaluate the expression \\(\left(\frac{2^2x^2}{xy^3}\
ight)^2\\) for \\(x = 4\\) and \\(y = 2\\). the solution is

Explanation:

Step1: Simplify the expression inside the parentheses

First, simplify \(\frac{2^{2}x^{2}}{xy^{3}}\). We know that \(2^{2}=4\), and using the rule of exponents \(\frac{a^{m}}{a^{n}} = a^{m - n}\) for \(a
eq0\), for the \(x\) terms: \(\frac{x^{2}}{x}=x^{2-1}=x\). So the expression inside the parentheses becomes \(\frac{4x}{y^{3}}\).

Step2: Square the simplified expression

Now, square \(\frac{4x}{y^{3}}\), using the rule \((\frac{a}{b})^{n}=\frac{a^{n}}{b^{n}}\), we get \((\frac{4x}{y^{3}})^{2}=\frac{(4x)^{2}}{(y^{3})^{2}}\). Then, using \((ab)^{n}=a^{n}b^{n}\) for the numerator and \((a^{m})^{n}=a^{mn}\) for the denominator: \(\frac{16x^{2}}{y^{6}}\).

Step3: Substitute \(x = 4\) and \(y = 2\)

Substitute \(x = 4\) and \(y = 2\) into \(\frac{16x^{2}}{y^{6}}\). First, calculate \(x^{2}=4^{2}=16\) and \(y^{6}=2^{6}=64\). Then the numerator is \(16\times16 = 256\) and the denominator is \(64\). So \(\frac{256}{64}=4\).

Answer:

\(4\)